Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Quantum mechanics - spin operator eigenvalue probability?

  1. May 23, 2010 #1
    Hi,
    I have this problem on a past exam paper I am having some trouble with:

    "in the conventional basis of the eigenstates of the Sz operator, the spin state of a spin-1/2 particle is described by the vector:

    [tex] u = \left( \stackrel{cos a}{e^i^b sina} \right)[/tex] where a and B are constants.

    find the probability that a measurement of the y-component of the spin of the particle will yield the result [tex] 0.5\hbar [/tex] ."

    For the life of me I cannot work out how to write out matrices legibly on this thing, so I will summarize what is bothering me. I am given pauli matrices [tex] \sigma_x_,_y_,_z [/tex]that I cannot write out properly, and the spin operator is given by [tex] S_i = i\hbar\sigma_i [/tex].

    In the question I am given the vector u, which is apparently expressed in the basis of Sz eigenstates.

    Am I justified in putting this vector u into an eigenvalue equation [tex] S_{y} u = a_{y} u [/tex] ,

    where ay is my eigenvalue, when the vector I would be operating on is made from a basis of eigenstates of another operator (Sz)?

    I tried this and got two equations for ay, neither of which gives [tex] a_y = 0.5\hbar [/tex].

    does this mean I can conclude that there is zero probability of finding the y-component of the spin being equal to [tex] 0.5\hbar [/tex] ?

    or do I somehow have to wangle it so that I get another vector (not u) that is in the Sy eigenstate basis?

    thanks.



    PS. sorry, this crazy thing will not let me change something 5 lines up where I should have said
    "Am I justified in putting this vector u into an eigenvalue equation S_y u = 0.5\hbar u"
     
    Last edited: May 23, 2010
  2. jcsd
  3. May 23, 2010 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    You tell me. If [itex]u[/itex] is an eigenvector of [itex]S_z[/itex], is it also an eigenvector of S_y[/itex]?
     
  4. May 23, 2010 #3
    are you actually asking me, do these two operators commute? and if their commutator [Sy, Sz] = 0, then yes they do?

    is that right, if the eigenvector of one operator is equal to the eigenvector of another, then the operators commute?
     
  5. May 23, 2010 #4

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Yes. So, do [itex]S_z[/itex] and [itex]S_y[/itex] commute?
     
  6. May 23, 2010 #5
    not by my calculations, although I just woke up 5 minutes ago. I get [Sy,Sz] is equal to
    (hbar^2)/4 multiplied by a 2x2 matrix containing zeros on the leading diagonal and 2i elsewhere.
     
  7. May 23, 2010 #6

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Right, they don't commute; [itex][S_y,S_z]=i\hbar S_x[/itex]. So, [itex]u[/itex] is clearly not an eigenstate of [itex]S_y[/itex].

    What are the eigenstates of [itex]S_y[/itex]?
     
  8. May 23, 2010 #7
    i got that one of them was the column vector, say, x:
    x_1

    ix_1

    and the other vector y:
    y_1

    -iy_1

    but I was not convinced by this because, if the upper component is say, x1, and I choose x1 = 1, then if I go to normalize it i do
    1/sqrt[(1^2) + (i^2)] = 1/(1-1) = 0, so how can this be right if Im getting a normalization constant of zero?
     
  9. May 23, 2010 #8

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You forgot to take the conjugate. You should be calculating [itex]\langle x|x \rangle = x^\dagger x[/itex], not [itex]x^T x[/itex].
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook